Question

If $\left(5\right)^{2}\left(5\right)^{4}\left(5\right)^{6}\ldots .\left(5\right)^{2 n}=\left(0.04\right)^{- 28}$ , then the value of $n$ is equal to

• A. $7$
• B. $5$
• C. $6$
• D. $3$

Answer 1

Answer: (A)

The given equation can be written as $5^{2}\left(5^{2}\right)^{2}\left(5^{2}\right)^{3}\ldots \left(5^{2}\right)^{n}=\left(5^{2}\right)^{28}$
$\left(5^{2}\right)^{1 + 2 + 3 + \ldots + n}=\left(5^{2}\right)^{28}$
$\Rightarrow \frac{n \left(n + 1\right)}{2}=28$
$\Rightarrow n=7$